Questions

# A calorimeter of water equivalent 20 g contains 180g of water at ${25}^{0}$C. ‘m’ grams of steam at ${100}^{0}$C is mixed in it till the temperature of the mixture is ${31}^{0}$C. The value of ‘m’ is close to (Latent heat of water = 540 cal ${g}^{-1}$, specific heat of water = 1 cal ${g}^{-1}$  )

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a
2
b
4
c
3.2
d
2.6
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detailed solution

Correct option is A

Heat gained by cold body = heat lost by hot body⇒20+180×1×31-25=m(540)+m×1×100-31⇒1200=m(609)⇒m≈2gms

Similar Questions

A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40oC. When m gram of ice at -10oC is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20oC. It is known that specific heat capacity of the liquid changes with temperature as where $\mathrm{\theta }$ is temperature in oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are  and latent heat of fusion of ice is ${\mathrm{L}}_{\mathrm{f}}=80{\mathrm{calg}}^{-1}$. Assume no heat loss to the surrounding and calculate the value of m in grams.

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